If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field,then the orbit will be or The path executed by a charged particle whose motion is perpendicular to magnetic field is

$A$ charged particle with a velocity $2 \times 10^{3} \ ms^{-1}$ passes undeflected through electric and magnetic fields in mutually perpendicular directions. The magnetic field is $1.5 \ T$. The magnitude of the electric field will be:

When an electron is accelerated by a potential difference of $V$ volts and enters a magnetic field,the force acting on it is $F$. What will be the force acting on the electron when it is accelerated by a potential difference of $5V$ volts and enters the same magnetic field?

$A$ charged particle is moving in a circular orbit of radius $6 \, cm$ with a uniform speed of $3 \times 10^6 \, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4} \, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

An $\alpha$-particle travels in a circular path of radius $0.45\,m$ in a magnetic field $B = 1.2\,Wb/m^2$ with a speed of $2.6 \times 10^7\,m/s$. The period of revolution of the $\alpha$-particle is: